Binomial sequences Andrzej Nowicki Toru´ n 30.09.2017 Contents 1 Introduction 1 2 Notations and preliminary facts 2 3 Initial properties of principal sequences 4 4 Principal power series 6 5 Properties of binomial sequences 10 6 Linear operators of type zero 11 7 Examples of binomial sequences 17 7.1 Successive powers of x ........................... 17 7.2 Lower and upper factorials ......................... 18 7.3 Abel polynomials .............................. 19 7.4 Laguerre polynomials ............................ 23 7.5 Other examples ............................... 23 1 Introduction Throughout this article K is a field of characteristic zero, K [x] is the ring of poly- nomials in one variable x over K , and K [x, y] is the ring of polynomials in two variables x, y over K . Moreover, K [x][[t]] is the ring of formal power series in one variable t over K [x]. Let F =(F n (x)) n>0 be a nonzero sequence of polynomials in K [x]. We say that F is a binomial sequence if F n (x + y)= n X k=0 n k F k (x) F n-k (y) for all n > 0. The equalities are in the ring K [x, y]. The assumption that F is nonzero means that there exists a nonnegative integer n such that F n (x) 6= 0. We will say that a binomial sequence F =(F n (x)) n>0 is strict if every polynomial F n (x) is nonzero. The well known binomial theorem can be stated by saying that (x n ) n>0 is a strict binomial sequence. Several other such strict sequences exist. The sequence of lower factorials ( x (n) ) n>0 , defined by x (0) = 1 and x (n) = x(x-1)(x-2) ··· (x-n+1) for n > 1, is a strict binomial sequence. The same property has the sequence of upper factorials
Throughout this article K is a field of characteristic zero, K[x] is the ring of poly-nomials in one variable x over K, and K[x, y] is the ring of polynomials in two variablesx, y over K. Moreover, K[x][[t]] is the ring of formal power series in one variable t overK[x].
Let F = (Fn(x))n>0 be a nonzero sequence of polynomials in K[x]. We say that Fis a binomial sequence if
Fn(x+ y) =n∑k=0
(n
k
)Fk(x)Fn−k(y)
for all n > 0. The equalities are in the ring K[x, y]. The assumption that F is nonzeromeans that there exists a nonnegative integer n such that Fn(x) 6= 0. We will say thata binomial sequence F = (Fn(x))n>0 is strict if every polynomial Fn(x) is nonzero.
The well known binomial theorem can be stated by saying that (xn)n>0 is a strictbinomial sequence. Several other such strict sequences exist. The sequence of lowerfactorials
(x(n))n>0
, defined by x(0) = 1 and x(n) = x(x−1)(x−2) · · · (x−n+1) for n >1, is a strict binomial sequence. The same property has the sequence of upper factorials
Andrzej Nowicki, 2017, Binomial sequences 2
(x(n))n>0
, defined by x(0) = 1 and x(n) = x(x + 1)(x + 2) · · · (x + n − 1) for n > 1.
The sequence of Abel’s polynomials (An(x))n>0, defined by A0(x) = 1 and An(x) =x(x − n)n−1 for n > 1, is a strict binomial sequence (see Subsection 7.3). Manyinteresting results concerning binomial sequences can be find for example in [13], [3],[20], [21], [23], [5], [15], and others.
There exists a full description of all strict binomial sequences. The important roleof such description play results of I. M. Sheffer [24], on linear operators of type zero,published in 1939. Later, in 1957, H. L. Krall [13], applying these results, proved thatF = (Fn(x))n>0 is a strict binomial sequence if and only if there exists a formal power
series H(t) =∞∑n=1
antn, belonging to K[[t]] with a1 6= 0 and without the constant term,
such that∞∑n=0
Fn(x)
n!tn = exH(t).
In Section 6 we present his proof and some basic properties of linear operators of typezero. Several other proofs and applications of this result can be find; see for example:[21], [20] and [12]. We have here the assumption that F is strict. In the known proofsthis assumption is very important. In this case every polynomial Fn(x) is nonzero andmoreover, degFn(x) = n for all n > 0.
However there exist non-strict binomial sequences. We have the obvious exampleF = (1, 0, 0, . . . ). The sequence (Fn(x))n>0 defined by
F2m(x) =(2m)!
m!xm and F2m+1(x) = 0 for all m > 0,
is also a non-strict binomial sequence. Some other such examples are in Section 7.
In this article we present a description of all binomial sequences. We prove (seeTheorem 5.5) that if in the above mentioned result of Krall [13] we omit the assumptiona1 6= 0, then this result is also valid for non-strict binomial sequences.
2 Notations and preliminary facts
We denote by N the set {1, 2, 3, . . . }, of all natural numbers, and by N0 the set{0, 1, 2, . . . } = N ∪ {0}, of all nonnegative integers.
If i1, . . . , is ∈ N0, where s > 1, then we denote by⟨i1, . . . , is
⟩the generalized
Newton integer(i1 + · · ·+ is)!
i1! · · · is!.
In particular, 〈i1〉 = 1, 〈i1, i2〉 =(i1+i2i1
),⟨i1, i2, i3
⟩=⟨i1 + i2, i3
⟩⟨i1, i2
⟩.
Let F = (Fn)n>0 be a nonzero sequence of polynomials belonging to K[x]. Let usrepeat that F is a binomial sequence if
Fn(x+ y) =∑i+j=n
〈i, j〉Fi(x)Fj(y) for all n > 0.
Andrzej Nowicki, 2017, Binomial sequences 3
We shall say that F is a principal sequence, if
Fn(x+ y) =∑i+j=n
Fi(x)Fj(y) for all n > 0.
Here the sums range over all pairs of nonnegative integers (i, j) such that i + j = n.We say that a binomial sequence F is strict if all the polynomials Fn are nonzero.Moreover, we say that a principal sequence F is strict if all the polynomials Fn arenonzero.
Proposition 2.1. Let F = (Fn)n>0 and P = (Pn)n>0 be nonzero sequences of polyno-mials from K[x] such that
Pn =1
n!Fn for n > 0.
The sequence F is binomial if and only if the sequence P is principal. Moreover, F isa strict binomial sequence if and only if P is a strict principal sequence.
Proof. Assume that F is binomial. Then we have
Pn(x+ y) = 1n!Fn(x+ y) = 1
n!
∑i+j=n
〈i, j〉Fi(x)Fj(y) =∑
i+j=n
(1i!Fi(x)
) (1j!Fj(y)
)=
∑i+j=n
Pi(x)Pj(y).
Hence, it is clear that P is principal. The opposite implication is also clear. �
Thus, if we have a result for principal sequences, then by the above proposition weobtain a similar result for binomial sequences.
Let R be a commutative ring with identity. We shall denote by R<<t>> the ringof formal power series with divided powers ([2], [18]). Every its element is an ordinary
formal power series of the form∞∑n=0
rntn with rn ∈ R. It is the ring with the usual
addition and with the multiplication ∗ defined by the formulas a ∗ tn = tn ∗ a = atn fora ∈ R, and
tn ∗ tm = 〈n,m〉 tn+m =
(n+m
n
)tn+m.
This multiplication ∗ is called the binomial convolution ([10], [18]). If f =∞∑n=0
antn and
g =∞∑n=0
bntn are elements of R<<t>>, then the binomial convolution of f and g is
f ∗ g =∞∑n=0
(∑i+j=n
〈i, j〉aibj
)tn.
The ring R<<t>> is commutative with identity equals 1. Note that if f =∞∑n=0
antn,
g =∞∑n=0
bntn and h =
∞∑n=0
cntn, then
(f ∗ g) ∗ h = f ∗ (g ∗ h) =∞∑n=0
( ∑i+j+k=n
〈i, j, k〉aibjck
)tn.
If R is a domain containing Q, then R<<t>> is also a domain.
Andrzej Nowicki, 2017, Binomial sequences 4
Proposition 2.2. If Q ⊂ R, then the rings R<<t>> and R[[t]] are isomorphic. Moreexactly, the mapping σ : R<<t>> → R[[t]] defined by
σ
(∞∑n=0
f(n)tn
)=∞∑n=0
f(n)
n!tn
is an isomorphism of rings.
Proof. It is clear that σ is a bijection, σ(1) = 1 and σ(F +G) = σ(F ) + σ(G) for
F,G ∈ R<<t>>. Put F =∞∑n=0
f(n)tn and G =∞∑n=0
g(n)tn. Then F ·G =∞∑n=0
(f ∗g)(n)tn
and we have
σ(FG) =∞∑n=0
1n!
(f ∗ g)(n)tn =∞∑n=0
1n!
( ∑i+j=n
〈i, j〉f(i)g(j)
)tn
=∞∑n=0
( ∑i+j=n
(f(i)i!
)(g(j)j!
))tn =
(∞∑n=0
f(n)n!tn)(
∞∑n=0
g(n)n!tn)
= σ(F )σ(G).
This completes the proof. �
3 Initial properties of principal sequences
Proposition 3.1. If P = (Pn)n>0 is a principal sequence, then P0 = 1.
Proof. Suppose P0 = 0. Then P1(x) = P1(x + 0) = P0(x)P1(0) + P1(x)P0(x) =0 + 0 = 0. Let n > 2 and assume that P0 = P1 = · · · = Pn−1 = 0. Then
Hence, by induction, Pn = 0. Thus, if P0 = 0 then P is the zero sequence; but it is acontradiction because by definition every principal sequence is nonzero.
Therefore, P0 6= 0. Let P0 = pnxn+pn−1x
n−1+· · ·+p0, where n > 0, p0, . . . , pn ∈ K,and pn 6= 0. Since P0(x+ x) = P0(x)P0(x), we have the equality
2npnxn + 2n−1pn−1x
n−1 + · · ·+ 2p1x+ p0 = p2nx2n + · · ·+ p20.
If n > 1, then p2n = 0 but this contradicts the assumption pn 6= 0. Thus, n = 0 andP0 = p0 ∈ K r {0}. Moreover, p0 = p20, because P0(0) = P0(0 + 0) = P0(0)2. Hence,P0 = p0 = 1. �
Proposition 3.2. If P = (Pn)n>0 is a principal sequence, then Pn(0) = 0 for all n > 1.
Proof. We already know from Proposition 3.1 that P0 = 1. This implies thatP1(0) = P1(0 + 0) = P1(0) + P1(0), so P1(0) = 0. Let n > 1 and assume thatP1(0) = P2(0) = · · · = Pn(0) = 0. Then
Pn+1(0) = Pn+1(0 + 0) =∑
i+j=n+1
Pi(0)Pj(0) = Pn+1(0) + Pn+1(0)
Andrzej Nowicki, 2017, Binomial sequences 5
and so, Pn+1(0) = 0. �
Assume that P = (Pn)n>0 is an arbitrary principal sequence. We do not assume thatP is strict. There exist many non-strict such sequences. For example P = (1, 0, 0, . . . )is a non-strict principal sequence. Next such examples we may obtain by the followingproposition.
Proposition 3.3. Let (Pn)n>0 be a principal sequence and let s be a positive integer.Let (Rn)n>0 be a sequence of polynomials defined by
Rms = Pm for m > 0,
and Rn = 0 when s - n. Then (Rn)n>0 is a non-strict principal sequence.
Proof. It is obvious that Rn(x + y) = 0 =∑
i+j=n
Gi(x)Gj(y) in the case when
s - n. If n = sm with m ∈ N0, then∑
i+j=sm
Ri(x)Rj(y) =∑
si+sj=bm
Rsi(x)Rsj(y) =∑i+j=m
Pi(x)Pj(y) = Pm(x+ y) = Rsm(x+ y). �
Note also the following general property of principal sequences.
Proposition 3.4. Let (Pn(x))n>0 be a principal sequence of polynomials from K[x] andlet 0 6= a ∈ K. Let
Rn(x) = anPn(x) for n > 0.
Then (Rn(x))n>0 is a principal sequence.
Proof. We have
Rn(x+ y) = anPn(x+ y) = an∑
i+j=n
Pi(x)Pj(y)
=∑
i+j=n
(aiPi(x)) (ajPj(y)) =∑
i+j=n
Ri(x)Rj(y).
This completes the proof. �
In the next proposition we characterize strict principal sequences.
Proposition 3.5. Let (Pn(x))n>0 be a strict principal sequence. Then:
(1) P1(x) = ax for some 0 6= a ∈ K;
(2) degPn(x) = n for all n > 0;
(3) the initial monomial of each Pn(x) is equal to 1n!anxn.
Proof. Let P1(x) = amxm+am−1x
m−1+ · · ·+a0, where m > 0 and a0, . . . , am ∈ Kwith am 6= 0. Since P1(x + y) = P0(x)P1(y) + P1(x)P0(y) and P0(x) = P0(y) = 1, wehave (putting y = x) P1(2x) = 2P1(x) and so, (2m − 2)am = 0. Hence, m = 1 becauseam 6= 0. We know also that F1(0) = 0 (see Proposition 3.2). Therefore,
P1(x) = ax for some 0 6= a ∈ K.
Andrzej Nowicki, 2017, Binomial sequences 6
Now let s > 2 and assume that the initial monomial of every Pk(x), for k = 1, . . . , s−1,is equal to 1
k!akxk. Look at the equality
Ps(2x)− 2Ps(x) =s−1∑k=1
Pk(x)Ps−k(x).
On the right side we have a polynomial and its initial monomial is equal to
s−1∑k=1
(1
k!akxk
)(1
(s− k)!as−kxs−k
)=
1
s!asxs
s−1∑k=1
(s
k
)=
2s − 2
s!asxs 6= 0.
This implies that Ps(x) 6= 0. Let Ps(x) = amxm + am−1x
m−1 + · · · + a0, where m > 0and a0, . . . , am ∈ K with am 6= 0. Then the initial monomial of Fs(2x) − 2Fs(x) isequal to (2m − 2) amx
m. Hence,
(2m − 2) amxm =
2s − 2
s!asxs
and hence, m = s and am = 1s!as. Therefore, degPs(x) = s and the initial monomial
of Ps(x) equals 1s!asxs. This completes the proof. �
Colorary 3.6. A principal sequence (Pn)n>0 is strict if and only if P1 6= 0.
4 Principal power series
In this section K[x][[t]] is the ring of formal power series over K[x] in one variable t.Every element of this ring is of the form
P (x) =∞∑n=0
Pn(x)tn,
where (Pn(x))n>0 is a sequence of polynomials belonging to K[x]. We shall say thatthe series P (x) is principal if (Pn(x))n>0 is a principal sequence.
Proposition 4.1. Let P (x) =∞∑n=0
Pn(x)tn ∈ K[x][[t]]. The series P (x) is principal if
and only if in the ring K[x, y][[t]] it satisfies the equality
P (x+ y) = P (x)P (y).
Proof. Assume that the series P (x) is principal. Then (Pn(x))n>0 is a principalsequence, and then
P (x+ y) =∞∑n=0
Pn(x+ y)tn =∞∑n=0
( ∑i+j=n
Pi(x)Pj(y)
)tn
=
(∞∑n=0
Pn(x)tn)(
∞∑n=0
Pn(y)tn)
= P (x)P (y).
Andrzej Nowicki, 2017, Binomial sequences 7
Thus if P (x) is principal, then it is clear that P (x + y) = P (x)P (y). The oppositeimplication is also clear. �
Let F =∞∑n=0
Fntn be a formal power series belonging to K[X][[t]], and let G =
∞∑n=1
Gntn ∈ K[x][[t]] be a formal power series without the constant term. Consider the
substitution
F (G) =∞∑n=0
Fn
(∞∑j=1
Gjtj
)tn.
Since G has no the constant term, F (G) is a formal power series belonging to K[x][[t]].
Let us use this substitution for the power series F = e =∞∑n=0
1n!tn and G = xH(t) where
H(t) =∞∑n=1
antn ∈ K[[t]]. Denote this substitution by P (x). Thus, we have
where each Pj(x) is a polynomial belonging to K[x]. Initial examples:
P0(x) = 1,
P1(x) = a1x,
P2(x) = 12a21x
2 + a2x,
P3(x) = 16a31x
3 + a1a2x2 + a3x,
P4(x) = 124a41x
4 + 12a21a2x
3 + a1a3x2 + 1
2a22x
2 + a4x,
P5(x) = 1120a51x
5 + 16a31a2x
4 + 12
(a21a3 + a1a22)x
3 + (a1a4 + a2a3)x2 + a5x.
Proposition 4.2. Let H(t) ∈ K[[t]] be a formal power series without the constantterm, and let
P (x) = exH(t).
Then P (x) is a formal power series belonging to K[x][[t]] and this series is principal.
Moreover, if P (x) =∞∑n=0
Pn(x)tn and H(t) =∞∑n=1
antn, then an = P ′n(0) for all n > 1,
where each P ′n(x) is the derivative of Pn(x).
Proof. Since H(t) is without the constant term, the substitution exH(t) is welldefined and it is really an element of K[x][[t]]. Moreover,
P (x+ y) = e(x+y)H(t) = exH(t)+yH(t) = exH(t)eyH(t) = P (x)P (y).
Hence, by Proposition 4.1, the series P (x) is principal.Now we use the derivation d
dxof the ring K[x][[t]], and we have
∞∑n=1
P ′n(x)tn = P ′(x) =(exH(t)
)′= H(t)exH(t).
Hence,∞∑n=1
P ′n(0)tn = H(t)e0 = H(t) =∞∑n=1
antn and hence, an = P ′n(0) for all n > 1. �
Andrzej Nowicki, 2017, Binomial sequences 8
Now we shall prove that every principal power series is of the above form exH(t),where H(t) ∈ K[[t]] is a power series without the constant term. Before our proof, letus recall some well known properties of formal power series.
Assume thatR is a commutative ring with identity containing the field Q, of rationalnumbers, and let R[[t]] be the ring of formal power series over R. Denote byM the idealtR[[t]], and let 1+M = {1 + f ; f ∈M}. Note thatM is the set of all power series fromR[[t]] without the constant terms, and 1 +M is the set of all power series from R[[t]]with constant terms equal to 1. We have two classical functions Log : 1 +M → Mand Exp :M→ 1 +M, defined by
Log (1 + ξ) = ξ − 12ξ2 + 1
3ξ3 − 1
4ξ4 + . . . =
∞∑n=1
(−1)n+1
nξn,
Exp (ξ) = 1 + ξ + 12!ξ2 + 1
3!ξ3 + . . . =
∞∑n=0
1n!ξn = eξ,
for all ξ ∈M. It is well known that Log (Exp (ξ)) = ξ and Exp (Log (1 + ξ)) = 1 + ξ,for all ξ ∈M. As a consequence of these facts we obtain
Lemma 4.3. With the above notations:
(1) if ξ, ν ∈M and eξ = eν, then ξ = ν;
(2) for every ξ ∈M there exists a unique ν ∈M such that eν = 1 + ξ.
Now let R be the polynomial ring K[x], where K is a field of characteristic zero.
Lemma 4.4. Let F (x) be a polynomial from K[x] such that
(x+ y)F (x+ y) = xF (x) + yF (y).
Then F (x) ∈ K.
Proof. Suppose that F (x) 6∈ K. Let degF (x) = n > 1, and let F (x) =anx
n + an−1xn−1 + · · ·+ a0, where a0, . . . , an ∈ K with an 6= 0. Putting y = x, we have
2xF (2x) = 2xF (x) and so, F (2x) = F (x). This implies that 2nan = an, so 2n = 1 andwe have the contradiction: 0 = n > 1. �
Now we are ready to prove the following main result of this section.
Theorem 4.5. Let P = (Pn(x))n>0 be a nonzero sequence of polynomials from K[x].Then P is a principal sequence if and only if there exists a formal power series H(t),belonging to K[[t]] and without the constant term, such that
∞∑n=0
Pn(x)tn = exH(t).
Proof. Put P (x) =∞∑n=0
Pn(x)tn. We already know (see Proposition 4.2) that if
P (x) = exH(t) where H(t) ∈ K[[t]] is without the constant term, then P is principal.
Now assume that P is principal. Since P is nonzero, we know by Proposition 3.1that P0(x) = 1. Thus, by Lemma 4.3(2), there exists a formal power series B(x) ∈
Andrzej Nowicki, 2017, Binomial sequences 9
K[x][[t]], without the constant term, such that P (x) = eB(x). Put B(x) =∞∑n=1
Bn(x)tn,
where each Bn(x) is a polynomial from K[x]. Observe that, by Proposition 3.2, we haveP (0) = 1. Hence, 1 = P (0) = eB(0) and hence, by Lemma 4.3(1), we have the equalityB(0) = 0. Therefore, Bn(0) = 0 for all n > 1. This implies that for every n > 1 there
exists a polynomial An(x) ∈ K[x] such that Bn(x) = xAn(x). Put A(x) =∞∑n=1
An(x)tn.
Then B(x) = xA(x), and we have
P (x) = exA(x),
where A(x) is a power series from K[x][[t]] without the constant term. Since P isprincipal, we know, by Proposition 4.1, that P (x+ y) = P (x)P (y). Hence
e(x+y)A(x+y) = P (x+ y) = P (x)P (y) = exA(x)eyA(y) = exA(x)+yA(y)
and hence, (x+ y)A(x+ y) = xA(x) + yA(y) (see Lemma 4.3(1)), that is,
∞∑n=1
((x+ y)An(x+ y)
)tn =
∞∑n=1
(xAn(x) + yAn(y)
)tn.
So (x+y)An(x+y) = xAn(x)+yAn(y) for all n > 1 and so, by Lemma 4.4, every An(x)
is a constant an belonging to K. Consequently A(x) =∞∑n=1
antn. Thus, P (x) = exH(t),
where H(t) =∞∑n=1
antn. This completes the proof. �
The next propositions are consequences of the above theorem.
Proposition 4.6. If A(x), B(x) ∈ K[x][[t]] are principal power series, then the productA(x)B(x) is a principal power series.
Proof. It follows from Theorem 4.5 that A(x) = exH1(t) and B(x) = exH2(t), whereH1(t), H2(2) are some formal power series from K[[t]] without the constant terms. ThenA(x)B(x) = exH(t), where H(t) = H1(t) + H2(t) is a formal power series from K[[t]]without the constant term. Hence, again by Theorem 4.5, the power series A(x)B(x)is principal. �
Proposition 4.7. Let P (x) =∞∑n=0
Pn(x)tn ∈ K[x][[t]] be a principal power series.
Then P (x) is invertible in K[x][[t]], and the inverse P (x)−1 is a principal power series.Moreover,
P (x)−1 =∞∑n=0
Pn(−x)tn.
Proof. It follows from Theorem 4.5 that P (x) = exH(t), where H(t) is a formalpower series from K[[t]] without the constant term. Then P (x)P (−x) = exH(t)e−xH(t) =e0 = 1, and hence P (x)−1 = P (−x) = ex(−H(t)), and, again by Theorem 4.5, the seriesP (x)−1 is principal. �
Thus, the set of all principal power series from K[x][[t]] is a subgroup of the multi-plicative group of the ring K[x][[t]].
Andrzej Nowicki, 2017, Binomial sequences 10
5 Properties of binomial sequences
In the previous sections we proved several essential properties of principal sequences.Let us recall (see Proposition 2.1) that a sequence of polynomials (Pn(x))n>0 is principal
if and only if(n!Pn(x)
)n>0
is a binomial sequence. The following propositions are
immediate consequences of Proposition 2.1 and the suitable propositions from Section3.
Proposition 5.1. Let F =(Fn(x)
)n>0
be a binomial sequence. Then:
(1) F0(x) = 1;
(2) Fn(0) = 0 for all n > 1.
(3) Let s be a positive integer, and let G = (Gn(x))n>0 be a sequence defined by
Gms(x) =(ms)!
m!Fm(x) for m > 0,
and Gn(x) = 0 when s - n. Then G is a binomial sequence.
(4) Let 0 6= a ∈ K. Let Gn(x) = anFn(x) for n > 0. Then (Gn(x))n>0 is a binomialsequence.
Proposition 5.2. If F = (Fn(x))n>0 is a strict binomial sequence, then
(1) F1(x) = ax for some 0 6= a ∈ K;
(2) degFn(x) = n for all n > 0;
(3) the initial monomial of each Fn(x) equals anxn.
Proof. Use Propositions 3.5 and 2.1. �
Colorary 5.3. A binomial sequence (Fn)n>0 is strict if and only if F1 6= 0.
Proposition 5.4. Let H(t) ∈ K[[t]] be a formal power series without the constantterm, and let
exH(t) =∞∑n=0
Fn(x)
n!tn.
Then (Fn(x))n>0 is a binomial sequence. Moreover, if H(t) =∞∑n=1
antn, then n!an =
F ′n(0) for all n > 1, where each F ′n(x) is the derivative of Fn(x).
The following theorem is the main result of this article. It is an extension of Krall’sresult [13] mentioned in Introduction.
Theorem 5.5. Let F = (Fn(x))n>0 be a nonzero sequence of polynomials from K[x].Then F is a binomial sequence if and only if there exists a formal power series H(t),belonging to K[x][[t]] and without the constant term, such that
∞∑n=0
Fn(x)
n!tn = exH(t).
Andrzej Nowicki, 2017, Binomial sequences 11
Proof. Use Theorem 4.5 and 2.1. �
Let us recall (see Section 2) that we denote by K[x]<<t>> the ring of formal power
series with divided powers over K[x]. If∞∑n=0
Fn(x)tn is a formal power series belonging
to K[x]<<t>>, then we shall say that this series is binomial if (Fn(x))n>0 is a binomialsequence.
The following propositions are immediate consequences of Proposition 2.1 and thesuitable facts from the previous section.
Proposition 5.6. If F,G ∈ K[x]<<t>> are binomial power series, then the binomialconvolution F ∗G is a binomial power series.
Proposition 5.7. Let F =∞∑n=0
Fn(x)tn be a formal power series belonging to K[x]<<
t>>. If F is binomial, then F is invertible in K[x]<<t>>, and the inverse F−1 is abinomial power series, and moreover
F−1 =∞∑n=0
Fn(−x)tn.
Proposition 5.8. The set of all binomial series from K[x]<<t>> is a subgroup of themultiplicative group of the ring K[x]<<t>>.
It follows from Theorem 5.5 that every binomial sequence (Fn(x))n>0 is uniquely
determined by the formula∞∑n=0
Fn(x)n!
tn = exH(t), where H(t) ∈ K[[t]] is a formal power
series without the constant term. Thus for every sequence H = (h1, h2, . . . ) of elementsof K we obtain the unique nonzero binomial sequence F = (Fn(x))n>0 defined by the
formula∞∑n=0
Fn(x)n!
tn = exH(t), where H(t) =∞∑n=1
hntn. In this case we shall say that F is
the binomial sequence determined by H(t).
Proposition 5.9. Let H(t) =∞∑n=1
hntn ∈ K[[t]], and let (Fn)n>0 be the binomial se-
quence determined by H(t). Let 0 6= a ∈ K. Then (anFn)n>0 is the binomial sequence
determined by H(at) =∞∑n=1
hnantn.
Proof. This proposition follows from Theorem 5.5 and Proposition 5.1(4). �
6 Linear operators of type zero
In this section we consider strict binomial sequences. We recall some importantresults of I. M. Sheffer [24] and H. L. Krall [13], mentioned in Introduction. Throughoutthis section we denote by d the ordinary derivative d
dx.
Assume that F is a polynomial belonging to K[x]. We know that dn(F ) = 0 for alln > degF . Moreover, dn(xn) = n! and
dn(xm) = m(m− 1) · · · (m− n+ 1)xm−n for n 6 m.
Andrzej Nowicki, 2017, Binomial sequences 12
Proposition 6.1. Let J : K[x]→ K[x] be a K-linear map. Then there exists a uniquesequence (Ln(x))n>0, of polynomials from K[x], such that
J(F ) =∞∑n=0
Ln(x)dn(F )
for every F ∈ K[x].
Proof. Put Fn = J(xn) for all n ∈ N0. We define the Ln(x) recurrently by therelation
Fn = J(xn) =n∑k=0
Lk(x) · n(n− 1) · · · (n− k + 1)xn−k,
for n > 0. That is,
L0 = F0,
L1 = F1 − xL0,
L2 = 12
(F2 − x2L0 − 2xL1) ,
L3 = 16
(F3 − x3L0 − 3x2L1 − 6xL2) ,
L4 = 124
(F4 − x4L0 − 4x3L1 − 12x2L2 − 24xL3) ,
and so on. Then, for every m ∈ N0, we have the equality
J(xm) =∞∑n=0
Ln(x)dn(xm).
But the mappings J and d areK-linear, hence J(F ) =∞∑n=0
Ln(x)dn(F ), for all F ∈ K[x].
It is obvious that such sequence (Ln(x))n>0 is unique. �
Thus, for every K-linear mapping J : K[x] → K[x] we have the unique sequence(Ln(x))n>0 associated with J . In this case the mapping J is said to be an operator oftype zero ([24], [13]) if its associated sequence is of the form: Ln(x) = cn ∈ K for alln > 0 with c0 = 0 and c1 6= 0, that is, if
J(F ) = c1d(F ) + c2d2(F ) + c3d
3(F ) + · · ·
for all F ∈ K[x], where cn ∈ K for n > 1 and c1 6= 0. There are many interestingpapers on operators of type zero, their generalizations and applications ([1], [25]).
Now we present some properties of operators of type zero.
Proposition 6.2 ([24]). Let J be an operator of type zero. If F ∈ K[x] is a nonzeropolynomial of degree n > 1, then J(F ) is a nonzero polynomial of degree n− 1.
Proof. Put J = c1d+ c2d2 + · · · , with c1 6= 0, and let F = anx
n + · · ·+ a1x+ a0,where a0, . . . , an ∈ K, an 6= 0. Then d(F ) = nanx
n−1 + · · · is a nonzero polynomial ofdegree n− 1, and the degrees of all the polynomials d2(F ), d3(F ), . . . are smaller thann− 1. Since c1 6= 0, the polynomial J(F ) = c1d(F ) + c2d
(F ) + · · · is nonzero, and itsdegree is equal to n− 1. �
Andrzej Nowicki, 2017, Binomial sequences 13
Proposition 6.3. Let J be an operator of type zero, and let G ∈ K[x] be a nonzeropolynomial of degree n − 1 > 0. Then there exists a unique polynomial F ∈ K[x] ofdegree n such that J(F ) = G and F (0) = 0.
Proof. Put J = c1d+ c2d2 + · · · , with c1 6= 0, and G = g0 + g1x+ · · ·+ gn−1x
n−1,where g0, . . . , gn−1 ∈ K, gn−1 6= 0. We shall construct a polynomial
F = f1x+ f2x2 + · · ·+ fnx
n
with f1, . . . , fn ∈ K and fn 6= 0, such that J(F ) = G.
If 1 6 j 6 m, the we use the notation:
w(m, j) = m(m− 1) · · · (m− j + 1).
Observe that, for all j−1, . . . , n, we have dj(F ) =n∑k=j
w(k, j)fkxk−j. If G = J(F ), then
we have the following equalities:
G =n∑j=1
cjdj(F ) =
n∑j=1
cjn∑k=j
w(k, j)fkxk−j
= c1
(w(1, 1)f1x
0 + w(2, 1)f2x1 + w(3, 1)f3x
2 + · · ·+ w(n, 1)fnxn−1)
+ c2
(w(2, 2)f2x
0 + w(3, 2)f3x1 + w(4, 2)f4x
2 + · · ·+ w(n, 2)fnxn−2)
+ c3
(w(3, 3)f3x
0 + w(4, 3)f4x1 + w(5, 3)f5x
2 + · · ·+ w(n, 3)fnxn−3)
...
+ cn−1
(w(n− 1, n− 1)fn−1x
0 + w(n, n− 1)fnx1)
+ cn
(w(n, n)fnx
0).
Comparing the coefficients of xn−1, we have gn−1 = c1w(n, 1)fn = nc1fn. But nc1 6= 0,so fn = 1
nc1gn−1. Thus, if J(F ) = G, then the coefficient fn uniquely determined. Now
compare the coefficients of xn−2. We have gn−2 = (n − 1)c1fn−1 + c2w(n, 2)fn. Butfn is already constructed and (n − 1)c1 6= 0, so the coefficient fn−1 is also uniquelydetermined.
Repeating this procedure we obtain the coefficients fn, fn−1, . . . , f2. In the finalstep, we compare the coefficients of x0 and we obtain the equality
g0 = c1f1 + (2!)c2f2 + · · ·+ (n!)cnfn.
But the coefficients f2, f3, . . . , fn are already uniquely determined and c1 6= 0, so thecoefficient f1 is also uniquely determined. This completes the proof. �
As a consequence of Proposition 6.3 we obtain
Proposition 6.4 ([24]). If J is an operator of type zero, then there exists a uniquesequence (Bn(x))n>0, of nonzero polynomials from K[x], such that:
(1) B0(x) = 1;
(2) Bn(0) = 0 for n > 1;
(3) J(Bn(x)) = Bn−1(x) for n > 0 where B−1(x) = 0.
Andrzej Nowicki, 2017, Binomial sequences 14
Proof. Put B0(x) = 1.Then of course J(B0(x)) = 0 = B−1(x). Let n > 0 andassume that the polynomials B0(x), B1(x), . . . , Bn(x) are already defined. Then, byProposition 6.3, there exists a unique nonzero polynomial Bn+1(x) ∈ K[x] such thatBn+1(0) = 0 and J (Bn+1(x)) = Bn(x). Thus, by induction, we obtain a uniquelydetermined sequence (Bn(x))n>0 satisfying the given conditions. �
The polynomial sequence (Bn(x))n>0 from the above proposition is said to be thebasic sequence of J (see [24], [13]). We will prove that this sequence is principal.
Let J = c1d+ c2d2 + . . . be a fixed operator of type zero. Let us recall that c1 6= 0
and cn ∈ K for n > 1. Denote by M(t) the formal power series from K[[t]], defined by
M(t) = c1t1 + c2t
2 + c3t3 + · · · .
Since M(t) is without the constant term and c1 6= 0. There exists a unique formalpower series
H(t) = s1t1 + s2t
2s3t3 + · · · ∈ K[[t]]
such that s1 = c−11 6= 0 and H(M(t)
)= M
(H(t)
)= t. Consider the formal power
series A(x) = exH(t). This series belongs to K[x][[t]]. Put
A(x) = exH(t) = A0(x) + A1(x)t+ A2(x)t2 + · · · ,
where An ∈ K[x] for all n > 0. It is clear that A0(x) = 1, An(0) = 0 for n > 1.Moreover, each An(x) is nonzero and degAn(x) = n.
Lemma 6.5 ([24]). If J and A are as above, then
J(An(x)
)= An−1(x) for n > 1.
Proof. Let us extend the derivative d = ddx
: K[x] → K[x] to the derivatived : K[x][[t]]→ K[x][[t]] putting d(t) = 0. Then
d
(∞∑n=0
fn(x)tn
)=∞∑n=0
d(fn(x)
)tn
and, for every k > 0, we have
dk
(∞∑n=0
fn(x)tn
)=∞∑n=0
dk(fn(x)
)tn.
Extend also the operator J : K[x]→ K[x] to the K[[t]]-linear mapping J : K[x][[t]]→K[x][[t]] defined by
J(ϕ) =∞∑n=1
cndn(ϕ),
for ϕ ∈ K[x][[t]. Since for every F ∈ K[x] there exists an m such that dm(F ) = 0, the
extended operator J is well defined. Observe that J
(∞∑p=0
Ap(x)tp)
=∞∑p=0
J(Ap(x)
)tp.
Andrzej Nowicki, 2017, Binomial sequences 15
In fact,
J
(∞∑p=0
Ap(x)tp)
=∞∑n=1
cndn
(∞∑p=0
Ap(x)tp)
=∞∑n=1
cn
(∞∑p=0
dn(Ap(x)
)tp)
=∞∑n=1
∞∑p=0
cndn(Ap(x)
)tp =
∞∑p=0
(∞∑n=1
cndn(Ap(x)
))tp
=∞∑p=0
J(Ap(x)
)tp.
Observe also that d(exH(t)
)= H(t)exH(t) and dk
(exH(t)
)= H(t)kexH(t) for all k > 0.
Hence,
J
(∞∑p=0
Ap(x)tp)
= J(exH(t)
)=∞∑n=1
cndn(exH(t)
)=∞∑n=1
cnH(t)nexH(t)
=
(∞∑n=1
cnH(t)n)exH(t) = M(H)exH(t) = texH(t)
= t
(∞∑p=0
Ap(x)tp)
=∞∑p=1
Ap−1(x)tp.
Hence, we proved that∞∑p=1
J(Ap(x)
)tp =
∞∑p=1
Ap−1(x)tp and this implies that J(An(x)
)=
An−1(x) for all n > 1. This completes the proof. �
Theorem 6.6 ([24]). If (Bn(x))n>0 is the basic sequence of an operator J =∑n=1
cndn
of type zero. then∞∑n=0
Bn(x)tn = exH(t),
where H(t) ∈ K[[t]] is the formal power series (without the constant term) such that
M(H) = H(M) = t, where M(t) =∞∑n=1
cntn.
Proof. Put exH(t) =∞∑n=0
An(x)tn. It is clear that A0(x) = 1 and An(0) = 0 for
n > 1. Moreover we know, by Lemma 6.5, that J (An(x)) = Jn−1(x) for all n > 0.Hence, by Proposition 6.4, the sequence (An(x))n>0 is the basic sequence of J . Thus,
Bn(x) = An(x) for n > 0, and we have the equality∞∑n=0
Bn(x)tn = exH(t). �
Theorem 6.7 ([24], [13]). The basic sequence of every operator of type zero is astrict principal sequence.
Proof. This is an immediate consequence of Theorem 6.6 and Proposition 4.2. �
Now we shall prove that every strict principal sequence is the basic sequence of anoperator of type zero. For this aim, first we prove two lemmas. Let us recall that K isa field of characteristic zero.
Andrzej Nowicki, 2017, Binomial sequences 16
Lemma 6.8. Let F (x), G(x) be two polynomials from K[x] such that
F (x+ y)− F (x)− F (y) = G(x+ y)−G(x)−G(y).
Then F (x) = G(x) + px for some p ∈ K.
Proof. Let F (x) = anxn+an−1x
n−1+ · · ·+a1x+a0 and G(x) = bnxn+bn−1x
n−1+· · · + b1x + b0, where a0, . . . , an, b0, . . . , bn ∈ K. We do not assume that an 6= 0 andbn 6= 0. Putting y = x, we have the equality F (2x)− 2F (x) = G(2x)− 2G(x), that is,
(2n − 2)anxn + (2n−1 − 2)an−1x
n−1 + · · ·+ 4a2x2 + a0
= (2n − 2)bnxn + (2n−1 − 2)bn−1x
n−1 + · · ·+ 4b2x2 + b0.
Observe that we have not the monomials a1x and b1x. This equality implies thataj = bj for j = 2, 3, . . . , n and a0 = b0. Thus, F (x) = G(x)+px where p = a1−b1 ∈ K.�
Lemma 6.9. Let (Pn)n>0 be a strict principal sequence. Then there exists a sequence(cn)n>1, of elements of K, such that c1 6= 0, and for every n > 1,
Vn
(Pj
)= Pj−1 for j = 1, 2, . . . , n,
where Vn = c1d+ c2d2 + · · ·+ cnd
n.
Proof. ([13]). We define the sequence (cn)n>1 recurrently by the following way.We know (see Proposition 3.5) that P1 = ax for some 0 6= a ∈ K, and the initialcoefficient of each polynomial Pn, for n > 1, is equal to 1
n!an.
Let c1 = 1a
and V1 = c1d. Then
V1
(P1
)=
1
ad(ax) =
a
a= 1 = P0.
Thus, c1 is determined. Let n > 2 and assume that the elements c1, . . . , cn−1 arealready determined. Consider the operator Vn−1 = c1d + c2d
2 + · · · + cn−1dn−1. We
already know that Vn−1 (Pj) = Pj−1 for j = 1, 2, . . . , n − 1. Since Vn−1 is an operatorof type zero, there exists the basic sequence (Bm)m>0 of Vn−1 (see Proposition 6.4). Itfollows from Proposition 6.3 that then Bj = Pj for all j = 0, 1, . . . , n − 1. Moreover,we know from Theorem 6.7 that (Bm)m>0 is a principal sequence. Hence,
Pn(x+ y)− Pn(x)− Pn(y) =n−1∑k=1
Pk(x)Pn−k(y) =n−1∑k=1
Bk(x)Bn−k(y)
= Bn(x+ y)−Bn(x)−Bn(y)
and hence, by Lemma 6.9, Pn = Bn+px for some p ∈ K. Moreover, sinceB1 = P1 = ax,the initial coefficient of Bn is equal to 1
n!an (see Proposition 3.5). We define
cn = − p
an+1.
Let Vn = c1d + · · · + cndn = Vn−1 + cnd
n. Then it is clear that Vn(Pj) = Pj−1 forall j = 1, 2, . . . , n − 1. We shall show that it is also true for j = n, that is, thatVn(Pn) = Pn−1. In fact,
Theorem 6.10 ([13]). Every strict principal sequence is the basic sequence of anoperator of type zero.
Proof. Let P = (Pn)n>0 be a strict principal sequence. Let (cn)n>1 be thesequence of elements from K, defined in Lemma 6.9. It follows from this lemma that
P is the basic sequence of the operator∞∑n=1
cndn. �
Now, by Proposition 2.1 and the above facts, we obtain
Theorem 6.11 ([13]). A sequence (Fn)n>0, of polynomials from K[x], is a strict
polynomial sequence if and only if(Fn
n!
)n>0
is the basic sequence of an operator of typezero.
We will say that (cn)n>1 is a strict sequence, if cn ∈ K for all n > 1 and c1 6= 0.Given an arbitrary strict sequence C = (cn)n>1, we obtain a unique strict binomialsequence (Fn)n>0 such that
(Fn
n!
)n>0
is the basic sequence of the operator
J = c1d+ c2d2 + c3d
3 + · · · .
We call it the C-sequence. Recall that d is the ordinary derivative ddx
. Every polynomialFn(x) is here nonzero, and its degree equals n. Moreover, every strict binomial sequenceis a C-sequence for some strict sequence C.
7 Examples of binomial sequences
7.1 Successive powers of x
It is well known that (xn)n>0 is a strict binomial sequence of polynomials. It isthe first classical example of binomial sequences. It is not difficult to verify that itis the C-sequence for C = (1, 0, 0, . . . ), and it is the binomial sequence determinedby H(t) = t. The binomial sequence (axn)n>0, where 0 6= a ∈ K, is determined byH(t) = at.
Example 7.1. Let F2n(x) = (2n)!n!xn and F2n+1(x) = 0 for all n > 0. Then (Fn(x))n>0
is the binomial sequence determined by H(t) = t2. This sequence is non-strict.
Let 0 6= a ∈ K and let s be a positive integer. Let F = (Fn(x))n>0, where
Fms(x) =(ms)!
m!anxn for m > 0,
and Fn(x) = 0 when s - n. Then F is the binomial sequence determined by H(t) =(at)s. If s > 2, then this sequence is non=strict.
Andrzej Nowicki, 2017, Binomial sequences 18
7.2 Lower and upper factorials
Let a ∈ K. Consider the polynomial sequence (Wn(x))n>0 defined by
Wn(x) =
{1, for n = 0,
x(x+ a
)(x+ 2a
)· · ·(x+ (n− 1)a
), for n > 1.
In particular, W1(x) = x, W2(x) = x2 + ax, W3(x) = x3 + 3ax2 + 2a2x, and
Wn+1(x) = (x+ na)Wn(x) for all n > 0.
Proposition 7.2. The sequence(Wn(x)
)n>0
is binomial.
Proof. We shall show, by induction, that for all n > 0,
Wn(x+ y) =∑i+j=n
〈i, j〉Wi(x)Wj(y)
It is obvious for n 6 1. Assume that it is true for some n > 1. Then Wn+1(x+ y) =
= (x+ y + na)Wn(x+ y) = (x+ y + na)n∑k=0
(nk
)Wk(x)Wn−k(y)
=n∑k=0
(nk
)(x+ ka)Wk(x)Wn−k(y) +
n∑k=0
(nk
)(y + (n− k)a
)Wk(x)Wn−k(y)
=n∑k=0
(nk
)Wk+1(x)Wn−k(y) +
n∑k=0
(nk
)Wk(x)Wn+1−k(y)
= Wn+1(x) +Wn+1(y) +n−1∑k=0
(nk
)Wk+1(x)Wn−k(y) +
n∑k=1
(nk
)Wk(x)Wn+1−k(y)
= Wn+1(x) +Wn+1(y) +n∑k=1
(nk−1
)Wk(x)Wn+1−k(y) +
n∑k=1
(nk
)Wk(x)Wn+1−k(y)
= Wn+1(x) +Wn+1(y) +n∑k=1
((nk−1
)+(nk
))Wk(x)Wn+1−k(y)
= Wn+1(x) +Wn+1(y) +n∑k=1
(n+1k
)Wk(x)Wn+1−k(y)
=n+1∑k=0
(n+1k
)Wk(x)Wn+1−k(y).
This completes the proof. �
Note that (Wn(x))n>0 is the binomial sequence determined by
We will show that this sequence is binomial. We will prove this fact trough a series oflemmas below. Let
Bn(x) =1
n!An(x) for n > 0.
Lemma 7.4. For every n > 1 and all 0 6 k 6 n− 1,
B(k)n (x) =
1
(n− k)!(x− ka)(x− na)n−1−k.
Here B(k)n (x) is the k-th derivative of Bn(x).
Proof. By induction on k. It is obvious for k = 0. Assume that it is true forsome k > 0. Then
B(k+1)n (x) =
(B
(k)n (x)
)′=(
1(n−k)!(x− ka)(x− na)n−1−k
)′= 1
(n−k)!
((x− na)n−1−k + (n− 1− k)(x− ka)(x− na)n−2−k
)= 1
(n−k)!(x− na)n−2−k(
(x− na) + (n− 1− k)(x− ka))
= 1(n−k)!(x− na)n−2−k(n− k)
(x− a(1 + k)
)= 1
(n−(k+1))!(x− na)n−1−(k+1)
(x− a(1 + k)
).
This completes the proof of this lemma. �
Andrzej Nowicki, 2017, Binomial sequences 20
Lemma 7.5.n∑p=1
(n
p
)pan−pzn−1 = n(z + a)n−1.
Proof. Use the derivative ddz
for the equalityn∑p=1
(np
)an−pzp = (z + a)n − 1. �
Lemma 7.6.n∑p=1
(n
p
)(z + pa)an−pzp−1 = (z + a+ an)(z + a)n−1 − an.
Proof. By Lemma 7.5, we have
n∑p=1
(np
)an−p(z + pa)zp−1 =
n∑p=1
(np
)an−pzp + a
n∑p=1
(np
)pan−pzp−1
= (z + a)n − an + an(z + a)n−1 = (z + a+ an)(z + n+ 1)(z + a)n−1 − 1.
This completes the proof. �
Lemma 7.7.n−1∑k=0
(n
k
)ak(x− (k + 1)a
)(x− (n+ 1)a
)n−1−k= x(x− an)n−1 − an.
Proof. Using Lemma 7.6 for z = x− (n+ 1)a, we obtain that the left side of the
above equality is equal ton−1∑k=0
(nk
)ak(z + (n− k)a
)zn−k−1 =
n∑p=1
(np
)an−p
(z + pa
)zp−1
= (z + a+ an)(z + a)n−1 − an = x(x− an)n−1 − an. �
Proposition 7.8. (An(x))n>0 is a strict binomial sequence. It is the C-sequence for
C =
(1, a,
1
2!a2,
1
3!a3,
1
4!a4, . . .
).
Proof. Put c1 = 1 and cn = 1(n−1)!a
n−1 for all n > 2, and let J = c1d+ c2d2 + · · · .
We need to show that J (Bn+1(x)) = Bn(x), that is, that
Bn(x) = c1B(1)n+1(x) + c2B
(2)n+1(x) + · · ·+ cn+1B
(n+1)n+1 (x)
for all n > 0. For n = 0 and n = 1 it is obvious. Assume that n > 2. Then, by theprevious lemmas, we have
J (Bn+1(x)) =∑k=1
B(k)n+1(x)
=n∑k=1
ak−1
(k−1)!(n−(k−1))!(x− ka)(x− (n+ 1)a
)n−k+ 1
n!an
=n−1∑k=0
ak
k!(n−k)!
(x− (k + 1)a
)(x− (n+ 1)a
)n−(k+1)
+ 1n!an
= 1n!
n−1∑k=0
(nk
)ak(x− (k + 1)a
)(x− (n+ 1)a
)n−(k+1)
+ 1n!an
= 1n!
(x(x− an)n−1 − an + an
)= 1
n!x(x− an)n−1 = Bn(x).
Andrzej Nowicki, 2017, Binomial sequences 21
This completes the proof. �
Thus, we already know that (An(x))n>0 is a binomial sequence. It is not difficultto check that this sequence is determined by
H(t) =∞∑n=1
(−na)n−1
n!tn.
The fact that (An(x))n>0 is a binomial sequence means that in the polynomial ring
K[x, y] we have the equalities An(x + y) =n∑k=0
(nk
)Ak(x)An−k(y) for all n > 0. Hence,
for a ∈ K and n > 0, the following identity holds
(1) (x+ y)(x+ y − na)n−1 =n∑k=0
(n
k
)x(x− ka
)k−1y(y − (n− k)a
)n−k−1.
Now we present a second proof of the above identity (1).
In 1826, Abel deduced an identity which is
(2) (x+ y)n =n∑k=0
(n
k
)x(x− ka)k−1(y + ka)n−k,
for a ∈ K. Many authors offered different proofs of this identity ([9], [7], [22], [8]).In 2004, M. Lipnowski [14] and G. Zheng [27] presented elegant and short proofsin Solutions of Problem 310 of Mathematical Olympiads’ Correspondence Program.There are many applications of the Abel identity ([7], [22], [11], [19], [26]).
Proposition 7.9. The identity (1) follows from the Abel identity.
Proof. Substitute in (2) the element −a to the places of a, and next substitutey + na to the places of y. Then we get
(3)n∑k=0
(n
k
)x(x+ ka
)k−1(y + (n− k)a
)n−k=(x+ y + na
)n.
Call Un(x, y, a) the left hand side of (3). Then Un(x, y, a) = (x+ y+na)n, and lookingat Un−1(x, y + a, a), we obtain the identity
(4)n−1∑k=0
(n− 1
k
)x(x+ ka
)k−1(y + (n− k)a
)n−1−k=(x+ y + na
)n−1.
Put P =n∑k=0
(nk
)x(x+ ka)k−1y
(y + (n− k)a
)n−k−1. Then we have
(x+ y + na)n =n∑k=0
(nk
)x(x+ ka
)k−1(y + (n− k)a
)n−k=
n∑k=0
(nk
)x(x+ ka
)k−1(y + (n− k)a
)n−1−k(y + (n− k)a
)= P +Q,
Andrzej Nowicki, 2017, Binomial sequences 22
where Q =n∑k=0
(nk
)x(x+ka
)k−1(n−k)a
(y+(n−k)a
)n−1−k. Using (4) and the identity
(n− k)(nk
)= n
(n−1k
)we get Q = na(x+ y + na)n−1. Hence, P = (x+ y + na)n −Q =
(x+ y + na)n − na(x+ y + na)n−1 = (x+ y + na)n−1(x+ y), and hence,
(x+ y)(x+ y + na)n−1 =n∑k=0
(n
k
)x(x+ ka)k−1y
(y + (n− k)a
)n−k−1.
Now, putting −a instead of a, we obtain (1). This completes the proof. �
Note also the following proposition.
Proposition 7.10. The Abel identity follows from the identity (1).
Proof. Substitute in (1) the element −a to the places of a, and next substitutey + na to the places of y. Then we get
(5) (x+ y + na)(x+ y)n−1 =n∑k=0
(n
k
)x(x− ka
)k−1(y + na)
(y + ka
)n−k−1.
We prove the Abel identity (2) by induction. When n = 0, then it is obvious. Assumethat for n > 1,
(6) (x+ y)n−1 =n−1∑k=0
(n− 1
k
)x(x− ka)k−1(y + ka)n−1−k.
Then, by (5), (6) and the identity n(n−1k
)= (n− k)
(nk
), we have
(x+ y)n = (x+ y + na)(x+ y)n−1 − na(x+ y)n−1
=n∑k=0
(nk
)x(x− ka
)k−1((y + ka) + (n− k)a
)(y + ka
)n−k−1−na
n−1∑k=0
(n−1k
)x(x− ka)k−1(y + ka)n−1−k
=n∑k=0
(nk
)x(x− ka
)k−1(y + ka
)n−k+na
n−1∑k=0
(n−1k
)x(x− ka)k−1(y + ka)n−1−k
−nan−1∑k=0
(n−1k
)x(x− ka)k−1(y + ka)n−1−k
=n∑k=0
(nk
)x(x− ka
)k−1(y + ka
)n−k.
This completes the proof. �
Andrzej Nowicki, 2017, Binomial sequences 23
7.4 Laguerre polynomials
Let (Ln(x))n>0 be the sequence of polynomials from K[x] defined by L0(x) = 1 and
Ln(x) =n∑k=1
n!
k!
(n− 1
k − 1
)xk, for n > 1.
They are called the Laguerre1 polynomials ([4], [6], [20], [12]). The first few polynomials:
[10] R. L. Graham, D. E. Knuth, O. Patashnik, Concrete Mathematics. A Foundationfor Computer Science, Adison-Wesley, 1989.
Andrzej Nowicki, 2017, Binomial sequences 26
[11] F. Huang and B. Liu, The Abel-type polynomials identities, The Electronic Journalof Combinatorics, 17 (2010) 1-7.
[12] V. V. Kisil, Polynomial sequences of binomial type and path integrals, Leeds-Math-Pure, Preprint 2001.
[13] H. L. Krall, Polynomials with the binomial property, American MathematicalMonthly 64(5) (1957) 342-343.
[14] M. Lipnowski, A solution of Problem 310, Mathematical Olympiads’ Correspon-dence Program, Canada, 2004.
[15] M. Mihoubi, Bell polynomials and binomial type sequences, Discrete Mathematics308 (2008) 2450-2459.
[16] W. Mlotkowski and A. Romanowicz, A family of sequences of binomial type,Preprint 2015.
[17] A. Nowicki, Arithmetic functions (in Polish), Podroze po Imperium Liczb, vol.5,Second Edition, Wydawnictwo OWSIiZ, Torun, Olsztyn, 2012.
[18] A. Nowicki, Factorials and binomial coefficiens (in Polish), Podroze po ImperiumLiczb, vol.11, Second Edition, Wydawnictwo OWSIiZ, Torun, Olsztyn, 2013.
[19] P. Petrullo, Outcomes of the Abel identity, Preprint, 2011.
[20] S. Roman and G.-C. Rota, The umbral calculus, Adv. in Math. 27 (1978) 95-188.
[21] G.-C. Rota, D. Kahaner, and A. Odlyzko, Finite operator calculus, J. Math. Anal.Appl. 42 (1973) 685-760.
[22] G.-C. Rota, J. Shen, B. D. Taylor, All polynomials of binomial type are representedby Abel polynomials, Ann. Sc. Norm. Super. Pisa Cl. Sci., 25 (1997), 731-738.
[23] J. Schneider, Polynomial sequences of binomial-type arising in graph theory,Preprint 2014.
[24] I. M. Sheffer, Some properties of polynomial sets of type zero, Duke. Math. J. 5(1939) 590-622.
[25] A. K. Shukla and S. J. Rapeli, An extension of Sheffer polynomials, ProyeccionesJournal of Mathematics 30(2) (2011) 265-275.
[26] S. Sykora, An Abel’s identity and its corollaries, Preprint 2014.
[27] G. Zheng, A solution of Problem 310, Mathematical Olympiads’ CorrespondenceProgram, Canada, 2004.
Nicolaus Copernicus University, Faculty of Mathematics and Computer Science,87-100 Torun, Poland, (e-mail: [email protected]).
